Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions

Briggs - Calculus: Early Transcendentals 3rd Edition

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Briggs 3rd Edition

Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions

Problem 7.2.44b

Chapter 7, Problem 7.2.44b

A running model A model for the startup of a runner in a short race results in the velocity function v(t) = a(1 - e⁻ᵗ/ᶜ), where a and c are positive constants, t is measured in seconds, and v has units of m/s. (Source: Joe Keller, A Theory of Competitive Running, Physics Today, 26, Sep 1973)

b. Using the velocity in part (a) and assuming s(0) = 0, find the position function s(t), for t ≥ 0.

Verified step by step guidance

Recall that the position function \(s(t)\) is the integral of the velocity function \(v(t)\) with respect to time \(t\). Mathematically, this is expressed as \(s(t) = \int v(t) \, dt + C\), where \(C\) is the constant of integration.

Substitute the given velocity function \(v(t) = a(1 - e^{-t/c})\) into the integral to get \(s(t) = \int a(1 - e^{-t/c}) \, dt + C\).

Split the integral into two parts: \(s(t) = a \int 1 \, dt - a \int e^{-t/c} \, dt + C\).

Integrate each term separately: the integral of 1 with respect to \(t\) is \(t\), and the integral of \(e^{-t/c}\) with respect to \(t\) involves using the substitution \(u = -t/c\), so \(\int e^{-t/c} \, dt = -c e^{-t/c} + D\), where \(D\) is another constant of integration.

Combine the results to write \(s(t) = a \left( t + c e^{-t/c} \right) + C\). Use the initial condition \(s(0) = 0\) to solve for \(C\) by substituting \(t=0\) into the expression and setting \(s(0) = 0\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Velocity and Position Relationship

Velocity is the rate of change of position with respect to time. Mathematically, velocity v(t) is the derivative of position s(t), so s(t) can be found by integrating v(t) over time. Given an initial position, integrating the velocity function yields the position function.

Integration of Exponential Functions

Integrating functions involving exponentials like e^(-t/c) requires understanding the antiderivative rules for exponential decay. The integral of e^(-kt) is (-1/k)e^(-kt) plus a constant. This is essential to find the position function from the given velocity.

Initial Conditions in Differential Equations

Initial conditions, such as s(0) = 0, are used to determine the constant of integration after integrating a velocity function. They ensure the position function accurately reflects the physical scenario, anchoring the solution to a specific starting point.

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